Problem: Evaluate the double integral. $ \int_1^{\ln(2)} \int_0^{e^x} -2x + 2y \, dy \, dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $2e^2 + 2e - 3 + \ln(2)$ (Choice B) B $2 - 3e$ (Choice C) C $6 - 4\ln(2) - \dfrac{e^2}{2}$ (Choice D) D $5 + 2\ln(2)$
First, we evaluate the inner integral. We can substitute in the $e^x$ at the end as if it were a numerical bound. $\begin{aligned} \int_1^{\ln(2)} \int_0^{e^x} -2x + 2y \, dy \, dx &= \int_1^{\ln(2)} \left[ -2xy + y^2 \right]_0^{e^x} dx \\ \\ &= \int_1^{\ln(2)} -2xe^x + (e^x)^2 dx \\ \\ &= \int_1^{\ln(2)} -2xe^x + e^{2x} dx \end{aligned}$ Second, we evaluate the outer integral. We can use integration by parts to find that the integral of $-2xe^x \, dx$ is $2(x - 1)e^x$. $\begin{aligned} &\int_1^{\ln(2)} -2xe^x + e^{2x} dx \\ \\ &= \left[ -2(x - 1)e^x + \dfrac{e^{2x}}{2} \right]_1^{\ln(2)} \\ \\ &= -2 (\ln(2) - 1)e^{\ln(2)} + \dfrac{e^{2\ln(2)}}{2} - \left( \dfrac{e^2}{2} \right) \\ \\ &= -4(\ln(2) - 1) + \dfrac{e^{\ln(4)}}{2} - \dfrac{e^2}{2} \\ \\ &= 6 - 4\ln(2) - \dfrac{e^2}{2} \end{aligned}$ The answer: $ \int_1^{\ln(2)} \int_0^{e^x} -2x + 2y \, dy \, dx = 6 - 4\ln(2) - \dfrac{e^2}{2}$